Question

(a) Let \(g\) be continuous on \((\alpha, \beta)\) and differentiable on the \(\left(\alpha, t_{0}\right)\) and \(\left(t_{0}, \beta\right) .\) Suppose \(A=\) \(\lim _{t \rightarrow t_{0}-} g^{\prime}(t)\) and \(B=\lim _{t \rightarrow t_{0}+} g^{\prime}(t)\) both exist. Use the mean value theorem to show that $$ \lim _{t \rightarrow t_{0}-} \frac{g(t)-g\left(t_{0}\right)}{t-t_{0}}=A \quad \text { and } \quad \lim _{t \rightarrow t_{0}+} \frac{g(t)-g\left(t_{0}\right)}{t-t_{0}}=B $$ (b) Conclude from (a) that \(g^{\prime}\left(t_{0}\right)\) exists and \(g^{\prime}\) is continuous at \(t_{0}\) if \(A=B\). (c) Conclude from (a) that if \(g\) is differentiable on \((\alpha, \beta)\) then \(g^{\prime}\) can't have a jump discontinuity on \((\alpha, \beta)\).

Short answer

Answer: If A = B, then the derivative g'(t_0) exists, and g' is continuous at t_0. Thus, if g is differentiable on \((\alpha, \beta)\), then g' cannot have a jump discontinuity on \((\alpha, \beta)\).

Step-by-step solution

Use the Mean Value Theorem for Left Limit

Let's first consider the left limit: $$\lim_{t\rightarrow t_0^-} \frac{g(t) - g(t_0)}{t - t_0} = A$$ Since \(g\) is continuous and differentiable on \((\alpha, t_0)\), we can apply the Mean Value Theorem on the interval \([t, t_0]\) (with \(t < t_0\)). It states that there exists a \(c_1\) in \((t, t_0)\) such that: $$g'(c_1) = \frac{g(t_0) - g(t)}{t_0 - t}$$ As \(t \rightarrow t_0^-\), \(c_1 \rightarrow t_0^-\). Hence, $$\lim_{t\rightarrow t_0^-} g'(c_1) = \lim_{t\rightarrow t_0^-} \frac{g(t) - g(t_0)}{t - t_0}$$ And since \(\lim_{t\rightarrow t_0^-} g'(c_1) = A\), we can conclude: $$\lim_{t\rightarrow t_0^-} \frac{g(t) - g(t_0)}{t - t_0} = A$$

Use the Mean Value Theorem for Right Limit

Now let's consider the right limit: $$\lim_{t\rightarrow t_0^+} \frac{g(t) - g(t_0)}{t - t_0} = B$$ This time, since \(g\) is continuous and differentiable on \((t_0, \beta)\), we can apply the Mean Value Theorem on the interval \([t_0, t]\) (with \(t > t_0\)). It states that there exists a \(c_2\) in \((t_0, t)\) such that: $$g'(c_2) = \frac{g(t) - g(t_0)}{t - t_0}$$ As \(t \rightarrow t_0^+\), \(c_2 \rightarrow t_0^+\). Hence, $$\lim_{t\rightarrow t_0^+} g'(c_2) = \lim_{t\rightarrow t_0^+} \frac{g(t) - g(t_0)}{t - t_0}$$ And since \(\lim_{t\rightarrow t_0^+} g'(c_2) = B\), we can conclude: $$\lim_{t\rightarrow t_0^+} \frac{g(t) - g(t_0)}{t - t_0} = B$$

Existence of g'(t0) and Continuity

If we have \(A = B\), then we can define the value of \(g'(t_0) = A = B\) according to the limits of the difference quotient. This also implies that g' has no jump discontinuity at \(t_0\). Therefore, if \(A = B\), then \(g'(t_0)\) exists, and g' is continuous at \(t_0\).

Conclude No Jump Discontinuity for g'

Part (a) tells us that if \(A = B\), then \(g'(t_0)\) exists, and g' is continuous at \(t_0\). Since this argument holds for an arbitrary point \(t_0\) in \((\alpha,\beta)\) and g is differentiable on \((\alpha, \beta)\), g' cannot have a jump discontinuity on \((\alpha, \beta)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In practice, these functions usually represent physical quantities, and their derivatives are a rate of change. When dealing with differential equations, understanding the underlying functions and their rates of change is crucial. For instance, in the original exercise, we focus on the function g and its derivative g'. We analyze the behavior of these functions close to a specific point t_0 to draw conclusions about continuity and differentiability. Differential equations are foundational in both pure and applied mathematics, as they describe various phenomena in fields like physics, engineering, and economics.

Continuity

The concept of continuity is a fundamental aspect of calculus. Continuity at a point means that the behavior of the function at that point is predictable and does not have any abrupt changes. For a function to be continuous at a point t_0, it must meet three criteria: the function must be defined at t_0, the limit as the function approaches t_0 must exist, and the function’s value at t_0 must equal this limit. The function g in the original exercise is stated to be continuous on the interval (alpha, beta), which sets the stage for applying the Mean Value Theorem—an essential tool for proving various properties about functions, like the existence of derivatives.

Differentiability

Differentiability refers to the ability to take a derivative at a point. If a function is differentiable at t_0, it means that its rate of change at t_0 can be described by a specific, well-defined slope—this is the derivative at that point. For a function to be differentiable at a point, it must also be continuous there, but the converse isn't always true. A differentiable function doesn't display an abrupt change in direction; the curve is smooth without any sharp corners. The step-by-step solution uses the concept of differentiability to apply the Mean Value Theorem, which allows us to equate the derivative of the function g within an interval to the function's average rate of change over that interval. This process is key to establishing the link between the function's derivatives and its continuity.

Limit of a Function

The limit of a function is a fundamental concept in calculus. It describes the value that the function approaches as the input (or the argument of the function) approaches a certain value. Limits are used to define both the continuity and the derivative of functions, supplying the foundation for the Mean Value Theorem. They also help in analyzing the behavior of functions near a point of interest, which is precisely what we see in the original exercise with the limits A and B. Understanding limits is crucial because they allow us to deal with values that are not explicitly calculable, such as when a function approaches an asymptote or when we need to check the function’s behavior near a point where it isn’t necessarily defined.